Let $B$ denote the closed unit ball in $\mathbf{R}^n$. Brouwer's fixed point theorem states that every continuous map $f:B\to B$ has a fixed point. There is a simple proof using Stokes's theorem, at least for the special case in which $f$ is smooth, as presented on Wikipedia here.
The page also states that this case contains the full generality of the theorem, because if $f:B\to B$ is continuous without fixed points then $\epsilon = \inf_{x\in B} |x-f(x)| > 0,ドル so we can just convolve (each component of) $f$ with a smooth bump $\psi:\mathbf{R}^n\to\mathbf{R}$ supported on $\epsilon B$ to get a smooth counterexample to the theorem.
Unfortunately, as it stands the proof doesn't work, because the distance of $f(B)$ to $\partial B$ could well be zero, in which case $\tilde{f} = \psi\ast f$ might not satisfy $\tilde{f}(B)\subset B$. Does anybody see a resolution to this difficulty?
EDIT, following Willy's answer. I've just realised that I was confused when I asked this question. $\tilde{f}(B)\subset B$ was never really an issue; the issue was rather that convolution isn't fully defined near the boundary. The most immediate interpretation is to extend $f:B\to B$ by 0ドル$ to $\mathbf{R}^n\to B,ドル but then mollifying $f$ doesn't give you a uniformly nearby $\tilde{f}$. The interpretation that works is to extend $f:B\to B$ to any uniformly continuous $F:\mathbf{R}^n\to B,ドル such as
$$F(x) = \begin{cases} f(x) & \text{if $|x|\leq 1,ドル}\\ f(x/|x|) & \text{if $|x|\geq 1,ドル}\end{cases}$$
and then mollify.
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2$\begingroup$ Why would you want to convolve? Isn't the point just that the smooth functions $f\colon B \to B$ are uniformly dense? $\endgroup$t.b.– t.b.2012年09月12日 08:58:59 +00:00Commented Sep 12, 2012 at 8:58
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1$\begingroup$ Yes, Wikipedia often contains irresponsible statements. In this case, see CS's comment which I have not verified myself. $\endgroup$Willie Wong– Willie Wong2012年09月12日 08:59:15 +00:00Commented Sep 12, 2012 at 8:59
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$\begingroup$ @t.b. I think Sean is quoting Wikipedia there... $\endgroup$Willie Wong– Willie Wong2012年09月12日 08:59:49 +00:00Commented Sep 12, 2012 at 8:59
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1$\begingroup$ I would just convolve in charts but the details are somewhat fiddly (I don't have the time to write this up now). Concerning the homotopy question, see this thread (I'm not sure if the manifolds with boundary case is included in those references, but Hirsch should do it). $\endgroup$t.b.– t.b.2012年09月12日 09:05:37 +00:00Commented Sep 12, 2012 at 9:05
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1$\begingroup$ Hmm, I suppose we could just convolve with a bump supported on $\epsilon B,ドル and then scale by 1ドル/(1+\epsilon)$. Points will move by at most 2ドル\epsilon$. $\endgroup$Sean Eberhard– Sean Eberhard2012年09月12日 09:07:51 +00:00Commented Sep 12, 2012 at 9:07
2 Answers 2
Sean's last comment inspired the following answer:
Let 100ドル\epsilon < \inf |x - f(x)|$. Let $g(x) = \frac{1}{1 + 10\epsilon} f(x)$. Then by triangle inequality we have that $|x - g(x)| > \epsilon/2$.
Let $h: (1+10\epsilon)^{-1}B \to (1+10\epsilon)^{-1}B$ be the smooth map formed by $$ h(x) = \eta* g(x) $$ where $\eta$ is a mollifier supported in $\epsilon B$. We have that $h(x)$ is smooth and has no fixed points etc.
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$\begingroup$ Neat trick!${}{}{}$ $\endgroup$t.b.– t.b.2012年09月12日 09:34:32 +00:00Commented Sep 12, 2012 at 9:34
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$\begingroup$ @t.b. Thanks. It occurred to me that uniform approximation is a bit stronger than we actually need (since we are arguing by contradiction). $\endgroup$Willie Wong– Willie Wong2012年09月12日 09:37:13 +00:00Commented Sep 12, 2012 at 9:37
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1$\begingroup$ Nice. If I might paraphrase, this argument can be described as follows. If convolving $f$ with a mollifier is saying "replace $f(x)$ with the its average (of sorts) on $x + \epsilon B$", then what you're saying is "replace $f(x)$ with its average on $x/(1+10\epsilon) + \epsilon B$". This is no longer a simple convolution, but it still replaces $f$ with a uniformly nearby smooth $\tilde{f}:B\to B$. $\endgroup$Sean Eberhard– Sean Eberhard2012年09月12日 09:44:56 +00:00Commented Sep 12, 2012 at 9:44
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1$\begingroup$ A similar idea, more closely related to my comment: Given $f:B\to B$ define $F:\mathbf{R}^n\to B$ by $F(x) = f(x)$ if $x\in B$ and $F(x) = f(x/|x|)$ otherwise. Now mollify $F$ and restrict the result to $B$. The result is a smooth $\tilde{f}:B\to B$ such that $\|f-\tilde{f}\|_\infty < \epsilon$. $\endgroup$Sean Eberhard– Sean Eberhard2012年09月12日 10:25:43 +00:00Commented Sep 12, 2012 at 10:25
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1$\begingroup$ @Sean: indeed. Furthermore, if $f$ has no fixed points, then clearly $F$ that you constructed also has no fixed points on $rB$ for any $r > 0$. In both of our constructions the trick is to get a map from a ball to a strictly smaller subset of itself. After which we can mollify and restrict. $\endgroup$Willie Wong– Willie Wong2012年09月12日 10:44:36 +00:00Commented Sep 12, 2012 at 10:44
hmm yes fix $\varepsilon$ greater than 0ドル$ and let $x_n$ be a convergent subsequence of a convergent sequence that converges to the same limit $l$. Then fix $\delta$ greater than, but not equal to 0ドル,ドル and let $|x_n-l|<\delta$ iff $f(x_n)\rightarrow f(l)$.
Furthermore I would suggest fixing $\varepsilon$ greater than 2ドル\delta/5$ and then let the convergent bijective map $f$ converge to a fixed limit $cl,ドル where $c$ is a non-linear constant.